Dx - k is Hypoelliptic

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Theorem

Let $f \in \map {C^\infty} \R$ be a smooth real function.

Let $T \in \map {\DD'} \R$ be a distribution.

Let $T_f$ be the distribution associated with $f$.

Let $k \in \R$.

Suppose in the distributional sense it holds that:

$\paren {\dfrac \d {\d x} - k} T = T_f \quad (1)$


Then there is a $c \in \R$ such that:

$T = T_{F + c \map \exp {k x} }$

where $F \in \map {C^\infty} \R$ is a solution to:



$\paren {\dfrac \d {\d x} - k} F = f$

In other words, $\dfrac \d {\d x} - k$ is a hypoelliptic operator.


Proof

By assumption $f \in \map {C^\infty} \R$.

By Solution to Linear First Order ODE with Constant Coefficients:

$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} F = f$

By Differentiable Function as Distribution and multiplication of distribution by a smooth function:

$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} T_F = T_f \quad (2)$

Subtract $(2)$ from $(1)$:

\(\ds \paren {\dfrac \d {\d x} - k} \paren {T - T_F}\) \(=\) \(\ds T_f - T_f\)
\(\ds \) \(=\) \(\ds \mathbf 0\)

where $\mathbf 0$ is the zero distribution.

This is solved by:

$T - T_F = T_{c \map \exp {k x} }$

or:

$T = T_{F + c \map \exp {k x} }$

where $c \in \C$.

Finally:

$\ds \paren {\dfrac \d {\d x} - k} \paren {F + c \map \exp {k x} } = f$

$\blacksquare$


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